The equation of the plane which passes through the point (3, 2, -1) and perpendicular to the line with direction ratios 2, 1, 2 is :

$2x+y+2z=6$

The correct answer is Option (3) → $2x+y+2z=6$

The equation of a plane passing through a given point $(x_0,y_0,z_0)$ and perpendicular to a given direction ratio is,

$a(x-x_0)+b(y-y_0)+c(z-z_0)=0$

where $(a,b,c)$ are direction ratios of the normal to the plane.

$∴2(x-3)+1(y-2)+2(z+1)=0$

$⇒2x+y+2z-6=0$

$⇒2x+y+2z=6$